Question

The sum of the series \[\dfrac{2}{3} + \dfrac{8}{9} + \dfrac{{26}}{{27}} + \dfrac{{80}}{{81}} + \]……… n terms are:
A) \[n - \dfrac{1}{2}({3^n} - 1)\]
B) \[n + \dfrac{1}{2}({3^n} - 1)\]
C) \[n - \dfrac{1}{2}(1 - {3^{ - n}})\]
D) \[n + \dfrac{1}{2}({3^{ - n}} - 1)\]

Answer 1

Hint: in this question we have to find n term of given series. Here we have to first find the pattern of the series. Rearrange the given series to and check for pattern. If any patterns is observed then follow the same pattern and use appropriate formula to get the required value.

Formula Used: We can find sum of n terms of GP by using
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
Where
\[{S_n}\]is sum of n terms of GP
a is first term of GP
n numbers of terms

Complete step by step solution: Given: \[\dfrac{2}{3} + \dfrac{8}{9} + \dfrac{{26}}{{27}} + \dfrac{{80}}{{81}} + \]
Now rearrange the above series
\[(1 - \dfrac{1}{3}) + (1 - \dfrac{1}{{{3^2}}}) + (1 - \dfrac{1}{{{3^3}}}) + ... + (1 - \dfrac{1}{{{3^n}}})\]
\[(1 + 1 + 1 + 1 + 1 + ..... + n) - (\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\]
\[(\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\] This series are in GP
\[n - (\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\]
Here \[a = \dfrac{1}{3}\]and \[r = \dfrac{1}{3}\]
We know that Sum of n terms of GP is given by
\[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
Where
\[{S_n}\]is sum of n terms of GP
a is first term of GP
n numbers of terms
\[n - (\dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + ..... + \dfrac{1}{{{3^n}}})\]
Now apply sum formula
\[ = n - \dfrac{{\dfrac{1}{3}\{ 1 - {{(\dfrac{1}{3})}^n}\} }}{{1 - (\dfrac{1}{3})}}\]
Now required value is
\[ = n + \dfrac{1}{2}({3^{ - n}} - 1)\]

Option ‘D’ is correct

Note: Whenever given series doesn’t follow any pattern then we first try to rearrange the series. If we get any pattern then follow that pattern to get required values. Sometimes by using pattern we are able to find the first term and common ratio therefore always try to find first term and common ratio.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.